If $x-y+z=0$, then what is the value of $(x^2+yz)÷(x^2+y^2+z^2)$?

Given the condition:
$$x-y+z=0⟹y=x+z$$
Let's assume simple numerical values that satisfy this condition to evaluate the expression efficiently:Let $x=1$ and $z=1$. Then:
$$y=1+1=2$$
Now, substitute $x=1$, $y=2$, and $z=1$ into the given algebraic expression:
$$\text{Numerator}=x^2+yz=1^2+(2\times 1)=1+2=3$$
$$\text{Denominator}=x^2+y^2+z^2=1^2+2^2+1^2=1+4+1=6$$
Dividing the two components:
$$\text{Value}=\frac{\text{Numerator}}{\text{Denominator}}=\frac{3}{6}=\frac{1}{2}$$